RAIRO. I NFORMATIQUE THÉORIQUE
P. G ORAL Č ÍK
Z. H EDRLÍN
V. KOUBEK
J. RYŠUNKOVÁ
A game of composing binary relations
RAIRO. Informatique théorique, tome 16, no 4 (1982), p. 365-369
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R.A.I.R.O. Informatique théorique/Theoretical Informaties
(vol. 16, n° 4, 1982, p. 365 à 369)
A GAME OF COMPOSING BINARY RELATIONS (*)
by P. GORALCÏK, Z. HEDRLIN,
V. KOUBEK et J. RYSUNKOVA (*)
Communicated by J. BERSTEL
Resumé. — Nous étudions la longueur du plus court produit de relations binaires, pris dans un
ensemble fini et donnant un résultat prescrit. On montre que dans certains cas, cette longueur croît
polynomialement en fonction de la taille des relations, et que dans d'autres cas, une telle majoration
riexiste pas.
Abstract. — We study the length of the shortest product of binary relations taken in a finite set
and giving a prescribed result. We show that in some cases this length grows polynomially with the
size of the relations, and that in other cases, such an upper bound does not exist.
The game is for one player, who is given a family M = {RU . . -,Rk} of
binary relations on a set X of size n and whose goal is to compose, as quickly
as possible, a relation S of a specified form from those in &. Of course, one
must be sure that it can be done, Le. that S belongs to the semigroup $+
generated by M. Next, since the only thinkable point in playing this sort of
patience is to kill time, it would be interesting to know how quickly - polynomially or not —the necessary time spent on it grows with the size n
of the underlying set X. The necessary time can be measured by the length of
the shortest séquence w = (HI-1, . . .,Ris) in St such that Rh. . Rts = S (where
on the left is the right-hand composition of w defined by: (x, z)eRil. . . Ris iff
there exists a séquence yo,yu • • -,ys such that yo = x, ys = z, and
fa-1,yu)eRtk
for ail k=\,
. . ,9s).
It will be convenient to write Rtl. . . Ris instead of (Ril9 . . ., Rt) when it is
clear if a word in 0t or the corresponding composition is meant.
Let us consider five types of possible resulting relation S : the identity
l x = {(x,x)|xeX}, a constant (i.e. a transformation of X with one-point
image), the universal relation X2, the empty relation Ç), a hyperconstant
(R<=X2 is a hyperconstant iff 3y Vx((x,y)eR)). We shall show that no one of
these can be composed in a polynomial time unless we confine the relations
(*). Received Mardi 1981, revised September 1981.
C) MFF KU, Sokolovska 83, 18600 Praha 8, Tchécoslovaquie.
R.A.I.R.O. Informatique théorique/Theoritical Informaties, 0399-0540/1982/365/S 5.00
© AFCET-Bordas-Dunod
366
p. GORALCÎK et al
occurring in M to some more restricted class # of relations, such as e. g.
partial transformations, (full) transformations, or rich relations (defined as the
relations with both projections equal to X).
Combining the classes <€ of relations just mentioned with the above five
possibilities for S we can formulate the following bunch of results :
1*
Constant
X2
<P Hyperconstant
General relations
N
N
N
N
N
Partial transformations
N
N
P
N
Transformations
N
P
Rich relations
N
P
P
P
In the above Table the filled entries assign to the corresponding
couples C^, S) true statements P or N = nonP9 where P = "there exists a
polynomial p (n) such that, for an arbitrary family ffi^ ofbinary relations on
an n-setX such that Seâ$+, the shortest séquence (Rit, . . .,Rt) in M with
Rix. . Ris = S has length s ^ ƒ>(>*)"•
THEOREM:
The blank entries in the Table indicate that 5 cannot be generally composed
from the relations in ^.
Proof: Thanks to the logical interconnections we soon discover between
some of the entries we need not prove all of them but concentrate upon the
key ones.
Starting with the first column, realize that 1* can be composed only from
permutations on X However, even in case 0t consists of a single permutation ƒ
we have no polynomial bound on the least positive integer r with fr= lx (the
k
order Ord( ƒ)). Indeed, taking n= £ ph the sum of the first k primes, and a
Ï=I
permutation ƒ with exactly one /?rcycle for each i= 1, . . ., fe, we have, using
the well-known Tchebyscheffs estimate [2], Pi^pk^k2
for i= 1, . . .,k, whence
k
n= Z Pi^kPk^k\
k
therefore Ord(f)=Hpt^kl^[^n]l9
,
which establishes
all the entries in the first column of the table.
We use the fact that there is no polynomial bound on the orders of
permutations for establishing all the other cases of non-polynomiality; ail of
R.A.I.R.O. Informatique théorique/Theoretical Informaties
A GAME OF COMPOSING BINARY RELATIONS
367
them will involve a fixed permutation ƒ of maximal order r on X and a fixed
transformation g taking each cycle of ƒ constantly into itself (see fig. 1).
Fîg. 1
The set Y=Xgf 1 = {yeX\yfelm(g)
= Xg}9 consisting of points marked
by little squares in figure 1, enters into description of the following relations:
R=YxX, the arrows go only from each point y e Y to ail points of X\
R = X2 — R, the complement of R;
\|/, a partial transformations of X with domain Y, taking Y as a constant
into itself.
(a) Let M = {f9g9R}. Then we have X2=gf ~1ReM+. The shortest word
in M giving X2 cannot start with R, for x R = Ç) if x $ Y, The fïrst occurrence
of R must be preceded by a subword gfk with Im(gfk)<=Y which is only
possible with ƒ*— ƒ " x . Thus the shortest word giving X2 is unique and equal
to gfr~1R, where r = Ord(/). This proves that X2 cannot be polynomially
composed from gênerai relations.
(b) Let & = {f,g9R}.
Then q)=gf~lRe^+,
The shortest word in M
giving Ç) must be ended by R. Throwing the last letter R away we must get
a relation with the image (second projection) contained in Y, which is possible
only if the last occurrence of R is preceded by gfr~x. For any we&+, either
wR = ({) or the second projection of wR is X Again, the shortest word in 0t
giving Ç) is unique and equal to gfr~lR, which proves that Ç) cannot be
polynomially composed from gênerai relations.
(c) Let M~ {/,g,x|/}. Then gf ~x \|/ is a constant contained in 01*. In the
shortest word in M giving a constant, \|/ cannot be the first letter. Moreover,
the fïrst occurrence of \|/ must be preceded by the shortest word in { ƒ g} giving
a transformation with image contained in 7, which is gfT~1. Thus gfr~x v|/ is
vol. 16, n°4, 1982
368
p. GORALCÎK et al
the shortest word giving a constant, which proves that neither a constant nor
a hyperconstant can be polynomially composed from partial transformations,
and thus neither from gênerai relations.
(d) Let M be a family of partial transformations such that Ç)e^£+.
Then the subset Z g l of ail points z such that some y e M is not defined in z
is non-void and for every x e X there exists a word w in M such that xw e Z,
thus xwcp = Ç) for a suitable (pe^. Let w = a ! . . . a k . If the séquence
xau xai a2, • . ., xai a2 • . . a* is not one-one then we clearly can find a shorter
word w' with xw' = xw5 thus we can assume that w is of length k^n— 1. Let
X= { x ls . . ., xn }. We can recursively define a séquence of words wi,. . . , wn
in M, each of length g n, such that:
Then w = w>i. . . wrt is a word of length ^ n2 giving Ç).
(e) This is well-known (c/. [1]). A simple argument is as follows: If M is a
set of transformations on X such that ^ + contains a constant then any pair
/ n\
of points of X can be merged by a word in 0t of length g ( 1 — 1 ând we
must merge at most n — 1 couples to get a constant, which yields the estimate
(ƒ) Let ^ be a set of rich relations on X such that 01* contains a
hyperconstant. Define a set of transformations ST by fe$~ iff there exists
Reffl such that ƒ ^R. Then we prove that ^~+ contains a constant: Let
X x { c } c R 1 ( . . Rfc5 where R,- are rich for ail j = 1, . . ., /c. Then we have a
matrix (xitJ)9 i= 1, . . ., n, 7 = 0, . . ., fc, such that {Xi,0, . • • > xn,0 } = ^>
xi, k = . . . =x„,k = c, and (x^-^x^OeR, for ail i = l , . . . ,n, j = 1, . . .,fe.
We form recursively a new matrix (yitj) of the same type as follows: we put
J ; i,j=xi j J forallj = 0, . . .,fc;toeach i=2, . . .,nwefind thesmallestj(O such
that Xi, j(i) =ySj jii} for some s < i (there must be such, since xt,fc = xStk for any i, s)
and put:
yij = Xij
for ail Ï = 2, . . .,n
and
ytj=ysj
for ail i = 2, . . .,n
and
where s is such that s<i and XiiJ(I) = x Si j (0 . Then:
is a partial transformation of X, (pjüK; for every j - 1, . . ., /c. Since /£,- is rich,
we can easily extend (p, to a full transformation ƒ, so that (pj^fj^Rj, for every
R.A.I.R.O. Informatique théoriqué/Theoretical Informaties
A GAME OF COMPOSING BINARY RELATIONS
359
7 = 1, . . . ,k. Clearly,/i. . .fk—Xx {c}. By (e), thisconstant can be composed
from 8T polynomially. Replacing in this composition each transformation by
a relation from 0t containing it, we get a polynomial composition of a
hyperconstant from the relations in M.
(g) Let X2e$+
for a set M of rich relations on X. Since X2 is a
hyperconstant, we have a word wi in 0t of length ^ n 3 which contains a
constant, say on cx e X Noting that converses of rich relations are again rich
and the converse of X2 is X2, we can, for the same reason, polynomially
compose a hyperconstant w^1, say on c2eX, from the converses of the
relations in M. Then vv2 e ^ + . There must exist a word uixvM not longer than
n— 1 such that ci u = c2. Then wt ww2 = ^ 2 polynomially.
The theorem is proved.
J.-E. Pin [1] asked about the length of a shortest word composing X2 from
a family of binary relations. Our resuit shows it is not polynomial. Likewise,
a full constant cannot be polynomially composed from partial functions (while
Pin shows that a partial constant can be so). Hence it makes différence if we
conceive synchronization in the incompletely specified automata as by a word
acting on the states as a full or only as a partial constant.
REFERENCES
1. J.-E. PIN, Le problème de la synchronisation et la conjecture de Cerny, in A. DELUCA,
Ed., Non Commutative Structures in Algebra and Geometrie Combinatories, C.N.R.,
1978, pp. 46-58.
2. W. SCHWARZ, Einfù'hrung in Methoden und Ergebnisse der Primzahltheorie, BIHochschultaschenbuch, 278/278a, Mannheim, 1969.
vol. 16, n° 4, 1982